MCQ
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = $a$) is
- A$\frac{a}{2}$
- B$a\sqrt 2 $
- ✓$\frac{a}{{\sqrt 2 }}$
- D$\frac{{a\sqrt 2 }}{3}$
✓
Answer
Correct option: C.
$\frac{a}{{\sqrt 2 }}$
c
(c) Suppose at displacement $y$ from mean position potential energy = kinetic energy
(c) Suppose at displacement $y$ from mean position potential energy = kinetic energy
==> $\frac{1}{2}m({a^2} - {y^2}){\omega ^2} = \frac{1}{2}m{\omega ^{\rm{2}}}{y^2}$
==> ${a^2} = 2{y^2}$
==> $y = \frac{a}{{\sqrt 2 }}$
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